Polynomial Equations of Higher Degree: Difficult Problems with Solutions

Solution:
Let [tex]f(x)=3x^4+4x^3-12x^2+1[/tex]; as a polynomial, [tex]f[/tex] is continuous on [tex]\mathbb{R}[/tex]. We compute
[tex]f(-3)=243-108-108+1=28\gt 0,\quad f(-1)=3-4-12+1=-12\lt 0,\quad f(0)=1\gt 0[/tex]
[tex]f(1)=3+4-12+1=-4\lt 0,\quad f(2)=48+32-48+1=33\gt 0[/tex]
So [tex]f[/tex] takes values of opposite signs at the endpoints of each of the intervals [tex]\left(-3, -1\right)[/tex], [tex]\left(-1, 0\right)[/tex], [tex]\left(0, 1\right)[/tex] and [tex]\left(1, 2\right)[/tex]. By Bolzano's theorem (the intermediate value theorem), [tex]f[/tex] has a zero in each of these intervals.
These intervals are pairwise disjoint, so the four zeros are different. A polynomial of degree [tex]4[/tex] has at most [tex]4[/tex] roots, so the equation has exactly four real roots, and they are distinct.

Solution:
This is a biquadratic equation. With [tex]t=x^{2}[/tex] it becomes
[tex]3t^{2}-2\left(2a^{2}+2b^{2}-ab\right)t+3a^{2}b^{2}=0[/tex]
For its discriminant we have
[tex]\frac{D}{4}=\left(2a^{2}+2b^{2}-ab\right)^{2}-(3ab)^{2}=\left(2a^{2}+2b^{2}-4ab\right)\left(2a^{2}+2b^{2}+2ab\right)[/tex]
[tex]\frac{D}{4}=2(a-b)^{2}\times 2\left(a^{2}+ab+b^{2}\right)=4(a-b)^{2}\left(a^{2}+ab+b^{2}\right)[/tex]
Since [tex]a\ne b[/tex] and [tex]a^{2}+ab+b^{2}\gt 0[/tex] (because [tex]a, b\gt 0[/tex]), we get [tex]D\gt 0[/tex], so the equation in [tex]t[/tex] has two distinct real roots [tex]t_{1}\ne t_{2}[/tex]. By Vieta's formulas
[tex]t_{1}t_{2}=a^{2}b^{2}\gt 0,\qquad t_{1}+t_{2}=\frac{2}{3}\left(2a^{2}+2b^{2}-ab\right)=\frac{2}{3}\left(2(a-b)^{2}+3ab\right)\gt 0[/tex]
The positive product shows that [tex]t_{1}[/tex] and [tex]t_{2}[/tex] have the same sign, and the positive sum shows that both are positive. Then
[tex]x^{2}=t_{1}\Rightarrow x=\pm\sqrt{t_{1}},\qquad x^{2}=t_{2}\Rightarrow x=\pm\sqrt{t_{2}}[/tex]
These four numbers are real and pairwise distinct, because [tex]t_{1}, t_{2}\gt 0[/tex] and [tex]t_{1}\ne t_{2}[/tex]. This proves the statement.

Solution:
Since [tex]n\gt 0[/tex], the function [tex]f(x)=x^{3}+nx-3n[/tex] is strictly increasing (it is a sum of the increasing functions [tex]x^{3}[/tex] and [tex]nx-3n[/tex]), so the equation has exactly one real root [tex]x_{0}[/tex]. Moreover
[tex]f(0)=-3n\lt 0,\qquad f(3)=27+3n-3n=27\gt 0\ \Rightarrow\ 0\lt x_{0}\lt 3[/tex]
Hence an integer root can only be [tex]x_{0}=1[/tex] or [tex]x_{0}=2[/tex].
For [tex]x_{0}=1[/tex]: [tex]1+n-3n=0[/tex] gives [tex]n=\frac{1}{2}[/tex], which is not an integer.
For [tex]x_{0}=2[/tex]: [tex]8+2n-3n=0[/tex] gives [tex]n=8[/tex]. Check:
[tex]x^{3}+8x-24=(x-2)\left(x^{2}+2x+12\right)[/tex]
so [tex]x=2[/tex] is indeed a root. The only value is [tex]n=8[/tex].

Solution:
Since [tex]-2[/tex] is a root,
[tex]-8+20-2a+b=0\Rightarrow b=2a-12[/tex]
Horner's scheme with [tex]-2[/tex] on the coefficients [tex]1, 5, a, 2a-12[/tex] gives [tex]1, 3, a-6[/tex] and remainder [tex](2a-12)-2(a-6)=0[/tex], hence
[tex]x^3+5x^2+ax+b=(x+2)\left(x^2+3x+a-6\right)[/tex]
The other two roots are the roots of [tex]x^2+3x+a-6=0[/tex]; they must be real, distinct and different from [tex]-2[/tex]. So the discriminant must be positive:
[tex]D=9-4(a-6)=33-4a\gt 0\Rightarrow a\lt \frac{33}{4}\Rightarrow a\le 8[/tex]
For [tex]a=8[/tex] the quadratic is [tex]x^2+3x+2=(x+1)(x+2)[/tex], so [tex]-2[/tex] is a double root and the equation has only two distinct roots; this value is rejected.
For [tex]a=7[/tex] (and [tex]b=2[/tex]) the equation [tex]x^2+3x+1=0[/tex] has the distinct real roots [tex]\frac{-3\pm\sqrt{5}}{2}[/tex], and [tex]-2[/tex] is not one of them, because [tex]4-6+1=-1\ne 0[/tex]. So the largest value is [tex]a=7[/tex].

Solution:
Let the roots be [tex]u[/tex], [tex]2u[/tex] and [tex]v[/tex]. Vieta's formulas give
[tex]u+2u+v=\frac{14}{24}\Rightarrow 3u+v=\frac{7}{12}[/tex]
[tex]u\times 2u+uv+2uv=-\frac{63}{24}\Rightarrow 2u^{2}+3uv=-\frac{21}{8}[/tex]
[tex]u\times 2u\times v=-\frac{45}{24}\Rightarrow 2u^{2}v=-\frac{15}{8}[/tex]
From the first equation [tex]v=\frac{7}{12}-3u[/tex]. Substituting into the second:
[tex]2u^{2}+\frac{7}{4}u-9u^{2}=-\frac{21}{8}\iff 8u^{2}-2u-3=0\iff u=\frac{3}{4}\text{ or }u=-\frac{1}{2}[/tex]
The third equation must also hold. For [tex]u=-\frac{1}{2}[/tex] we get [tex]v=\frac{25}{12}[/tex] and [tex]2u^{2}v=\frac{25}{24}\ne -\frac{15}{8}[/tex], so this case is rejected. For [tex]u=\frac{3}{4}[/tex]:
[tex]v=\frac{7}{12}-\frac{9}{4}=-\frac{5}{3},\qquad 2u^{2}v=2\times \frac{9}{16}\times \left(-\frac{5}{3}\right)=-\frac{15}{8}[/tex]
which is correct. So the roots are [tex]\frac{3}{4}[/tex], [tex]\frac{3}{2}[/tex] and [tex]-\frac{5}{3}[/tex]. Check:
[tex]24x^{3}-14x^{2}-63x+45=(4x-3)(2x-3)(3x+5)[/tex]

Solution:
Both factors contain [tex]2x^{2}+1[/tex], so put [tex]t=2x^{2}+1[/tex]. The equation becomes
[tex](t-3x)(t+5x)=9x^{2}\iff t^{2}+2xt-15x^{2}-9x^{2}=0\iff t^{2}+2xt-24x^{2}=0[/tex]
As a quadratic equation in [tex]t[/tex] it has the roots [tex]t=4x[/tex] and [tex]t=-6x[/tex], so
[tex](t-4x)(t+6x)=0[/tex]
Case [tex]1[/tex]: [tex]t=4x[/tex], i.e.
[tex]2x^{2}-4x+1=0\Rightarrow x=\frac{4\pm\sqrt{8}}{4}=\frac{2\pm\sqrt{2}}{2}[/tex]
Case [tex]2[/tex]: [tex]t=-6x[/tex], i.e.
[tex]2x^{2}+6x+1=0\Rightarrow x=\frac{-6\pm\sqrt{28}}{4}=\frac{-3\pm\sqrt{7}}{2}[/tex]
So the equation has four real roots: [tex]x_{1,2}=\frac{2\pm\sqrt{2}}{2}[/tex] and [tex]x_{3,4}=\frac{-3\pm\sqrt{7}}{2}[/tex].

Solution:
Put [tex]a=x^2+2x-3[/tex] and [tex]b=x^2-3x+2[/tex]. Then [tex]a+b=2x^2-x-1[/tex] and the equation reads
[tex]a^3+b^3=(a+b)^3\iff a^3+b^3=a^3+b^3+3ab(a+b)\iff 3ab(a+b)=0[/tex]
Factor each of the three expressions:
[tex]a=(x-1)(x+3),\qquad b=(x-1)(x-2),\qquad a+b=(x-1)(2x+1)[/tex]
so the equation becomes
[tex]3(x-1)^3(x+3)(x-2)(2x+1)=0[/tex]
Its roots are [tex]x=1[/tex] (a triple root), [tex]x=-3[/tex], [tex]x=2[/tex] and [tex]x=-\frac{1}{2}[/tex].

Solution:
Complete the square: [tex]f(x)=(x+5)^{2}-5[/tex], i.e. [tex]f(x)+5=(x+5)^{2}[/tex] for every [tex]x[/tex]. Applying this identity with [tex]f(x)[/tex] in place of [tex]x[/tex], and then repeating, we get
[tex]f(f(x))+5=\left(f(x)+5\right)^{2}=(x+5)^{4},\qquad f(f(f(x)))+5=(x+5)^{8}[/tex]
[tex]f(f(f(f(x))))+5=(x+5)^{16},\qquad f(f(f(f(f(x)))))+5=(x+5)^{32}[/tex]
So the equation is equivalent to [tex](x+5)^{32}=5[/tex]. The exponent is even, so its real solutions are
[tex]x+5=\pm\sqrt[32]{5}\Rightarrow x_{1,2}=-5\pm\sqrt[32]{5}[/tex]

Solution:
The numbers [tex]x-1[/tex] and [tex]x+3[/tex] are symmetric about their mean [tex]x+1[/tex], so put [tex]y=x+1[/tex]. Then [tex]x-1=y-2[/tex], [tex]x+3=y+2[/tex] and the equation becomes
[tex](y-2)^5+(y+2)^5=242y[/tex]
By the binomial theorem
[tex](y\pm 2)^5=y^5\pm 10y^4+40y^3\pm 80y^2+80y\pm 32[/tex]
so in the sum the terms with odd powers of [tex]2[/tex] cancel and the left side equals [tex]2y^5+80y^3+160y[/tex]. Hence
[tex]2y^5+80y^3-82y=0\iff 2y\left(y^4+40y^2-41\right)=0\iff 2y\left(y^2-1\right)\left(y^2+41\right)=0[/tex]
Since [tex]y^2+41\gt 0[/tex], we get [tex]y=0[/tex], [tex]y=1[/tex] or [tex]y=-1[/tex], and [tex]x=y-1[/tex] gives [tex]x=-1[/tex], [tex]x=0[/tex] or [tex]x=-2[/tex].

Solution:
Substitute [tex]x=\sqrt{2}[/tex], using [tex](\sqrt{2})^2=2[/tex], [tex](\sqrt{2})^3=2\sqrt{2}[/tex], [tex](\sqrt{2})^4=4[/tex], [tex](\sqrt{2})^5=4\sqrt{2}[/tex]:
[tex]4\sqrt{2}+20+2\sqrt{2}(a+10)+2(b+3a+10)+\sqrt{2}(2b+3a+10)+a+b+8=0[/tex]
[tex](7a+3b+48)+(5a+2b+34)\sqrt{2}=0[/tex]
Both brackets are rational numbers. If [tex]5a+2b+34\ne 0[/tex], then [tex]\sqrt{2}=-\frac{7a+3b+48}{5a+2b+34}[/tex] would be rational, which is impossible. Hence
[tex]\begin{cases}5a+2b+34=0\\7a+3b+48=0\end{cases}[/tex]
Multiplying the first equation by [tex]3[/tex], the second by [tex]2[/tex] and subtracting, we get [tex]a+6=0[/tex], so [tex]a=-6[/tex] and [tex]b=-2[/tex]. The equation becomes
[tex]x^5+5x^4+4x^3-10x^2-12x=0\iff x\left(x^4+5x^3+4x^2-10x-12\right)=0[/tex]
Since [tex]\pm\sqrt{2}[/tex] should be roots, divide the fourth-degree factor by [tex]x^2-2[/tex]; the division is exact:
[tex]x^4+5x^3+4x^2-10x-12=\left(x^2-2\right)\left(x^2+5x+6\right)=\left(x^2-2\right)(x+2)(x+3)[/tex]
So the equation is [tex]x(x-\sqrt{2})(x+\sqrt{2})(x+2)(x+3)=0[/tex] and its roots are [tex]-3[/tex], [tex]-2[/tex], [tex]-\sqrt{2}[/tex], [tex]0[/tex] and [tex]\sqrt{2}[/tex].

Solution:
We find the remainder of the division by [tex]x^{2}+kx+1[/tex]. Modulo the divisor we may replace [tex]x^{2}[/tex] by [tex]-kx-1[/tex]:
[tex]x^{3}=x\times x^{2}\equiv -kx^{2}-x\equiv k(kx+1)-x=\left(k^{2}-1\right)x+k[/tex]
[tex]x^{6}=\left(x^{3}\right)^{2}\equiv\left(k^{2}-1\right)^{2}x^{2}+2k\left(k^{2}-1\right)x+k^{2}\equiv\left(-k^{5}+4k^{3}-3k\right)x-k^{4}+3k^{2}-1[/tex]
Also [tex]-15x^{3}\equiv-15\left(k^{2}-1\right)x-15k[/tex] and [tex]-8x^{2}\equiv 8kx+8[/tex]. Adding everything (together with the free term [tex]2[/tex]), we get the remainder
[tex]R(x)=\left(-k^{5}+4k^{3}-15k^{2}+5k+15\right)x+\left(-k^{4}+3k^{2}-15k+9\right)[/tex]
The polynomial is divisible exactly when both coefficients are zero: [tex]A=-k^{5}+4k^{3}-15k^{2}+5k+15=0[/tex] and [tex]B=-k^{4}+3k^{2}-15k+9=0[/tex]. These two conditions are equivalent to [tex]B=0[/tex] and [tex]A-kB=0[/tex], where
[tex]A-kB=k^{3}-4k+15=(k+3)\left(k^{2}-3k+5\right)[/tex]
The quadratic factor has discriminant [tex]9-20\lt 0[/tex], so [tex]A-kB=0[/tex] only for [tex]k=-3[/tex]. Check in [tex]B[/tex]: [tex]-81+27+45+9=0[/tex], so both conditions hold and [tex]k=-3[/tex]. Indeed,
[tex]x^{6}-15x^{3}-8x^{2}+2=\left(x^{2}-3x+1\right)\left(x^{4}+3x^{3}+8x^{2}+6x+2\right)[/tex]